To find $\frac{dy}{dx}$ by implicit differentiation for the equation:

$\tan^{-1}(3x^2y) = x + 5xy^2$

we will differentiate both sides of the equation with respect to $x$.

Step 1: Differentiate the left-hand side

The left-hand side is $\tan^{-1}(3x^2y)$. We'll use the chain rule to differentiate it:

$\frac{d}{dx}\left[\tan^{-1}(3x^2y)\right] = \frac{1}{1 + (3x^2y)^2} \cdot \frac{d}{dx}(3x^2y)$

Next, differentiate $3x^2y$:

$\frac{d}{dx}(3x^2y) = 3 \cdot (2x \cdot y + x^2 \cdot \frac{dy}{dx}) = 6xy + 3x^2 \frac{dy}{dx}$

So the derivative of the left-hand side becomes:

$\frac{1}{1 + (3x^2y)^2} \cdot (6xy + 3x^2 \frac{dy}{dx})$

Step 2: Differentiate the right-hand side

The right-hand side is $x + 5xy^2$. Differentiating term by term:

1. $\frac{d}{dx}(x) = 1$
2. $\frac{d}{dx}(5xy^2) = 5 \cdot (y^2 + 2xy \cdot \frac{dy}{dx}) = 5y^2 + 10xy \frac{dy}{dx}$

So, the derivative of the right-hand side is:

$1 + 5y^2 + 10xy \frac{dy}{dx}$

Step 3: Set the derivatives equal

We now have:

$\frac{1}{1 + (3x^2y)^2} \cdot (6xy + 3x^2 \frac{dy}{dx}) = 1 + 5y^2 + 10xy \frac{dy}{dx}$

Step 4: Solve for $\frac{dy}{dx}$

Distribute the left-hand side:

$\frac{6xy + 3x^2 \frac{dy}{dx}}{1 + (3x^2y)^2} = 1 + 5y^2 + 10xy \frac{dy}{dx}$

Multiply both sides by $1 + (3x^2y)^2$ to eliminate the denominator:

$6xy + 3x^2 \frac{dy}{dx} = (1 + 5y^2)(1 + (3x^2y)^2) + 10xy \frac{dy}{dx} (1 + (3x^2y)^2)$

Rearrange the equation to isolate $\frac{dy}{dx}$:

$3x^2 \frac{dy}{dx} - 10xy \frac{dy}{dx} (1 + (3x^2y)^2) = (1 + 5y^2)(1 + (3x^2y)^2) - 6xy$

$\frac{dy}{dx} \left[3x^2 - 10xy (1 + (3x^2y)^2)\right] = (1 + 5y^2)(1 + (3x^2y)^2) - 6xy$

$\frac{dy}{dx} = \frac{(1 + 5y^2)(1 + (3x^2y)^2) - 6xy}{3x^2 - 10xy (1 + (3x^2y)^2)}$

This is the expression for $\frac{dy}{dx}$.

Would you like further details or explanations on any part of this process?

Related Questions:

1. How do you apply implicit differentiation to trigonometric functions?
2. What is the chain rule and how does it work in differentiation?
3. How do you simplify expressions involving derivatives after applying implicit differentiation?
4. Can you use implicit differentiation for higher-order derivatives?
5. How does the inverse trigonometric function differentiation formula work?

Tip:

When differentiating implicitly, always remember to treat $y$ as a function of $x$ and apply the chain rule accordingly when $y$ appears in terms multiplied by $x$.